Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.
The Correct Answer and Explanation is:
To find the acid dissociation constant KaK_aKa of a monoprotic acid, we use the information provided:
- Concentration of acid solution, [HA] = 0.0192 M
- pH = 2.53
Step 1: Find the [H+][H^+][H+] concentration
Use the pH equation: pH=−log[H+]\text{pH} = -\log[H^+]pH=−log[H+] [H+]=10−pH=10−2.53≈2.95×10−3 M[H^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}[H+]=10−pH=10−2.53≈2.95×10−3M
Step 2: Set up the ICE table for a monoprotic acid
HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-HA⇌H++A−
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HA | 0.0192 | −x-x−x | 0.0192−x0.0192 – x0.0192−x |
| H⁺ | 0 | +x+x+x | x=[H+]=2.95×10−3x = [H^+] = 2.95 \times 10^{-3}x=[H+]=2.95×10−3 |
| A⁻ | 0 | +x+x+x | x=2.95×10−3x = 2.95 \times 10^{-3}x=2.95×10−3 |
Step 3: Plug into the Ka expression
Ka=[H+][A−][HA]=x2[HA]−xK_a = \frac{[H^+][A^-]}{[HA]} = \frac{x^2}{[HA] – x}Ka=[HA][H+][A−]=[HA]−xx2 Ka=(2.95×10−3)20.0192−2.95×10−3=8.70×10−60.01625K_a = \frac{(2.95 \times 10^{-3})^2}{0.0192 – 2.95 \times 10^{-3}} = \frac{8.70 \times 10^{-6}}{0.01625}Ka=0.0192−2.95×10−3(2.95×10−3)2=0.016258.70×10−6 Ka≈5.35×10−4K_a \approx 5.35 \times 10^{-4}Ka≈5.35×10−4
✅ Final Answer:
Ka=5.35×10−4\boxed{K_a = 5.35 \times 10^{-4}}Ka=5.35×10−4
💡 Explanation (300 words)
In acid-base chemistry, the acid dissociation constant KaK_aKa measures the strength of a weak acid in water. For a monoprotic acid, which donates only one proton (H⁺) per molecule, the dissociation follows: HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-HA⇌H++A−
The extent to which this acid dissociates determines the concentration of hydrogen ions and, hence, the pH of the solution. In this problem, we were given the pH (2.53) and the initial concentration of the acid (0.0192 M), but not directly the KaK_aKa.
To find it, we first calculated the hydrogen ion concentration using the inverse log of the pH. This yielded [H+]=2.95×10−3 M[H^+] = 2.95 \times 10^{-3} \, \text{M}[H+]=2.95×10−3M. Assuming all of this came from the dissociation of HA, we said that the amount of HA dissociated is also x=[H+]x = [H^+]x=[H+].
Then, using the expression for Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}Ka=[HA][H+][A−], we substituted the values from the ICE table. Since HA loses a proton to form H⁺ and A⁻ in equal amounts, we set both product concentrations equal to xxx. The remaining undissociated HA was the initial amount minus xxx.
Solving the equation gave us Ka=5.35×10−4K_a = 5.35 \times 10^{-4}Ka=5.35×10−4, confirming that the acid is weak (since strong acids fully dissociate and do not have a measurable Ka in dilute aqueous solution).